Logarithm and Lubin-Tate formal group Let $K$ be a finite extension, by Milne's online note "class field theory", $m_{\mathbb{C}_p}$ has a natural $O_K$ module structure where the action is given by $[a]_f$. For such a $f$, there exists a unique formal group $F_f$ admitting $f$ as a endmorphism.
By wiki, the logarithm is a power series such that $\log(F(x,y))=\log(x)+\log(y)$. A pity is that Milne doesn't introduce the logarithm of a formal group.
Then the question is how to prove the following formula:
$$\log([a]_f(t))=a\log(t),$$where $a\in m_{\mathbb{C}_p}$ and $a\in O_K$.
The motivation is the proof of lemma 8 in this paper aiming at proving the existence of Lubin-Tate periods.
Thanks!
 A: It was only five months later that I noticed this question. I’m sorry, but I find @user760870’s comment deficient, maybe even wrong.
Perhaps we might define a prelogarithm of the formal group $F$ defined over the field $K$ to be a $K$-series $g(x)$ such that $g(F(x,y))=g(x)+g(y)$. Then you see that if $\lambda\in K$, $\lambda g$ is also a prelogarithm of $F$, indeed that the set of prelogarithms is a $K$-vector space: you can add two to get another.
If $g\ne0$ is a prelogarithm, then let’s see that $g'(0)\ne0$. Indeed, if $g(x)\equiv ax^m\pmod{x^{m+1}}$, i.e. if $ax^m$ is the first nonzero term of $g(x)$, then you get $g(F(x,y))\equiv a(x+y)^m\equiv ax^m+ay^m\pmod{x^{m+1}}$, impossible unless $m=1$ in characteristic zero. This is why we don’t talk about characteristic-$p$ logarithms. And this is why $\{\text{prelogs of $F$}\}\to K$ by the rule $g\mapsto g'(0)$is an injection, even a surjection if there is at least one nonzero prelogarithm. That is, the set of prelogs is at most one-dimensional as a $K$-space. If one-dimensional, the unique prelog with $g'(0)=1$ is the logarithm of $F$, sometimes written $\log_F(x)$.
You do see immediately that if $[a]$ is an endomorphism of $F$, then $\log_F\circ[a]$ is a prelogarithm, and your question is answered.
It’s another issue to show that there is a nonzero prelogarithm. I prefer to do it by a degree-by-degree computation, but you may also take $F_1(0,x)$, that’s the derivative of $F$ with respect to the left-hand variable, then set that variable to zero, and substitute $x$ for the remaining $y$. Then integrate the reciprocal of that, and magically you’ll get the logarithm.
